# Write the expression for the infinitesimal radial inertal force dfr look at pic

Write the expression for the infinitesimal radial inertal force dfr "look at pic"

Consider the non-uniform rigid disc of Figure 1. The density of the material is variable and depends
on the radial position: p = p (r). Its analytical expression is
p=po+p(1)
and where Po and P(1) are known and constants. Note that the density changes only with the radial coordinate:
for a given r changing θ does not affect the density. It should he observed that equation 1 implies that at
the center of the disk (r = O) the density presents its minimum value (p = Po) and for r = R the density
reaches the maximum value (p = po + P1 R)
The radius of the disc is R and its uniform thickness is h. The generalized coordinate is represented by
the angle θ (see Figure 1). The disc can only rotate with respect to G and the translation is not allowed. The
effects of the gravitational forces are not considered (the disc is supposed to rotate in a horizontal plane).
The angular acceleration dθ and the angular velocity d^2θ are supposed to be non-zero quantities. The generic infinitesimal element of disc is identified by the radial coordinate r (note that 0<r< R) and angular
position θ (note that O<θ< 2∏). The infinitesimal element of disc has mass dm and is subjected to the
infinitesimal radial inertial force dFr and the infinitesimal tangential inertial force dFθ

f=ma

## The Attempt at a Solution

f=ma >>> df(r)=dm*a(r)
a(r)= r"-r(θ')^2
dm=dr*dθ*h*d(rho)
∴ dfr=(-r(θ')^2)(dm)

is this correct based on the question, my English is not so good so I am not sure if i am answering what the problem is asking.
Also on the dm part I understand that infinitesimal is just a tiny piece of the disk so to find the mass i would need a change in radius a change in angle since the high states constant i leave h but since the mass changes i do d(rho)/dr since it changes with the radius but if i put it in my dm equation i would get something like this

dm=dr*dθ*h*d(rho)/dr >>>> dm=dθ*h*d(rho) ????? is this right?

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Simon Bridge
Homework Helper

dm=dr*dθ*h*d(rho)

##d\rho## ?

if you have ##\rho(r)## then ##dm = \rho(r)dV## (density times volume right?)
the volume element has height ##h##, length ##dr##, and width ##r.d\theta## so ##dV = h.r.dr.d\theta##

hmm yes i get most of your explanation but where I am confuse is, since the density changes with the radius i thought the way to write the infinitesimal part

dm=dρ*dV where dp implies that there is an infinitesimal change in density too
dV=h.r.dr.dθ

or should i write it your way

dm=ρ(r)dV Where it says that ρ is a function of the radius?
dV=h.r.dr.dθ

thank you

Simon Bridge
Homework Helper

dm=dρ*dV where dp implies that there is an infinitesimal change in density too
But there isn't one ... dm is the amount of mass in a volume element so small that the density is a constant inside it. One of the consequences of what "infinitesimal" means.

Your are constructing the integral by noting that at position (r,θ) there is an infinitesimal volume dV which contains a mass dm ... the same position has a volume mass-density of ρ(r,θ) (which you are given).

The other thing you needed to look at is how you worked out the volume element ... the arc length at radius r width dθ is rdθ.

There is a lot of freedom in constructing these ... you could decide to divide the disk into rings radius r and thickness dr ... then dV=2πr.dr ... or you could decide to integrate over the height as well, so dV=r.dr.dθ.dz ... and z is integrated from 0 to h. The choice is about how much of a shortcut you want to take at the start... often dictated by the geometry - for instance, in this case you probably wouldn't divide the disk into infinitesimal wedges (dV=R2h.dθ/2) since the density varies with radius, but if the density was a function of θ then that would make sense.